Except for those rare few souls that write N/C programs for the sheer joy of it, the ultimate purpose of owning, programming, or operating an N/C machine tool is to make a buck. The buck is made by performing a service or making a product better and at lower cost than the competetion can do it.
One of the major factors in machine tool economics is time. The longer it takes to do something, the more it costs. The length of time it takes to make a piecepart using N/C is a direct function of the feedrate used. The slower the feedrate, the longer it will take to run the piecepart. Other factors, such as making unnecessary cutting passes, are also important, but using the correct feedrate (and spindle speed) is perhaps the most important factor in the economics of N/C. The optimum feedrate is one that maximizes the Material Removal Rate (MRR) without stalling the spindle or breaking the cutter.
The classic method of determining the feedrate is to decide upon a chip load (how thick a chip you wish each tooth to peel off), typically from 0.003 to 0.020 inch. Then calculate what the spindle speed should be. Next, count the number of teeth on the cutter. Then multiply the three together:
Feedrate (IPM) = chipload (IN) * RPM * teeth
The results of that equation, while adequate for most manual machining operations, is often inadequate for high spindle horsepower N/C machines.
In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer.
A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors:
The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored.
For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM).
For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut.
For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM.
The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3.
| Table 3.3 Material Machining Factors (MMF) |
|
|---|---|
| Material | MMF |
| Aluminum | 0.12 |
| Brass | 0.4 |
| Low-carbon steel | 1.25 |
| High-carbon & low alloy steel | 1.5 - 1.9 |
| Stainless and tool steels | 1.8 - 2.2 |
| Superalloys | 2.5 - 5.0 |
For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle.
Since:
HP = MRR * MMF
and
MRR = depth of cut * width of cut * feedrate
it follows that:
HP = MMF * depth of cut * width of cut * feedrate.
Thus,
| Feedrate | = | HP --------------------------------------- MMF * depth of cut * width of cut |
In the previous example, if the spindle motor were 5 HP
| Feedrate | = | 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) |
= | 4 IPM |
| Per tooth chip load | = | feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth |
Assuming a spindle speed of 400 RPM and a 2-flute cutter,
| Chip load | = | 4.0 IPM --------------------------------------- 400 RPM * 2 teeth |
= | 0.005 inch per tooth |
Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000.
| Feedrate | = | 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) |
= | 41.7 IPM |
| Chip load | = | 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth |
= | 0.0052 inch per tooth |
| Next: Review Questions for Chapter 3 |
|---|
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Updated Jan. 9, 2002
Copyright © 1988-2002 by George Stanton and
Bill Hemphill
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